Auto-solve daily LeetCode problem using GPT-5-mini

Author: jeremymanning

problems/3212/gpt5-mini.md

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+# [Problem 3212: Count Submatrices With Equal Frequency of X and Y](https://leetcode.com/problems/count-submatrices-with-equal-frequency-of-x-and-y/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+We need to count submatrices that (1) contain grid[0][0], (2) have equal counts of 'X' and 'Y', and (3) contain at least one 'X'. "Contain grid[0][0]" means the submatrix must include the cell (0,0). For a submatrix (contiguous rows and columns) to include (0,0) and be valid indices, its top row and left column must be 0. So every valid submatrix is actually a prefix submatrix anchored at (0,0) with some bottom-right corner (i, j). That simplifies the problem: we only need to check the rectangle from (0,0) to (i,j) for all i, j.
+
+So for each cell (i,j) compute the counts of 'X' and 'Y' in the prefix rectangle (0,0)-(i,j). Count how many such prefixes have countX == countY and countX >= 1.
+
+We can compute prefix counts efficiently with a per-row running sum plus adding the previous-row prefix sums, keeping only O(n) space.
+
+## Refining the problem, round 2 thoughts
+- Because top-left is fixed to (0,0), there are m * n candidate submatrices only, so an O(m * n) solution is ideal and easy to achieve.
+- Compute prefix counts for 'X' and 'Y' for each (i, j) using running row sums plus previous-row prefix totals.
+- Edge cases:
+  - All dots ('.'): equal counts (0 == 0) but must have at least one 'X' — so answer should be 0.
+  - Very large grid up to 1000x1000: O(mn) time and O(n) extra space is fine.
+- No need for hashing or complicated transforms.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def countSubmatrices(self, grid: List[List[str]]) -> int:
+        """
+        Count prefix submatrices anchored at (0,0) whose counts of 'X' and 'Y' are equal
+        and contain at least one 'X'.
+        """
+        if not grid or not grid[0]:
+            return 0
+        m, n = len(grid), len(grid[0])
+        # prevX[j] and prevY[j] store the prefix sums for rows 0..i-1 and cols 0..j
+        prevX = [0] * n
+        prevY = [0] * n
+        ans = 0
+
+        for i in range(m):
+            runningX = 0
+            runningY = 0
+            prefX = [0] * n
+            prefY = [0] * n
+            for j in range(n):
+                ch = grid[i][j]
+                if ch == 'X':
+                    runningX += 1
+                elif ch == 'Y':
+                    runningY += 1
+                # prefix up to (i, j) = running sum in current row up to j + prefix of rows above up to j
+                prefX[j] = runningX + (prevX[j] if i > 0 else 0)
+                prefY[j] = runningY + (prevY[j] if i > 0 else 0)
+                # check conditions: equal counts and at least one X
+                if prefX[j] == prefY[j] and prefX[j] > 0:
+                    ans += 1
+            prevX = prefX
+            prevY = prefY
+
+        return ans
+```
+- Notes:
+  - We exploit that any submatrix containing (0,0) must be the prefix (0,0)-(i,j).
+  - For each row i we maintain runningX and runningY (row cumulative sums up to column j). Adding prevX[j] / prevY[j] (sums from earlier rows up to col j) yields full prefix sums.
+  - Time complexity: O(m * n) where m and n are grid dimensions.
+  - Space complexity: O(n) extra space (two arrays prevX and prevY of length n, and O(n) temporary arrays prefX/prefY per row).